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Q5(iii):
Evaluate the determinant: (iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$

Solution :

Step 1: Establishing the Determinant Structure

To evaluate the determinant of a $3 \times 3$ matrix $A$, we apply the Laplace expansion method. Let $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$. The determinant $|A|$ is calculated by expanding along any row or column. We shall expand along the first row ($R_1$).

The formula for expansion along $R_1$ is: $|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Step 2: Identifying Matrix Elements

Given the determinant $\Delta = \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$, we identify the coefficients:

  • $a_{11} = 0, a_{12} = 1, a_{13} = 2$
  • $a_{21} = -1, a_{22} = 0, a_{23} = -3$
  • $a_{31} = -2, a_{32} = 3, a_{33} = 0$

Step 3: Calculating Minors and Expansion

Substituting the values into the expansion formula [Per the Definition of Determinant of a Matrix]:

$\Delta = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}$

Calculating the $2 \times 2$ determinants:

  • First term: $0 \cdot [(0)(0) - (-3)(3)] = 0 \cdot [0 - (-9)] = 0$
  • Second term: $-1 \cdot [(-1)(0) - (-3)(-2)] = -1 \cdot [0 - (6)] = -1 \cdot (-6) = 6$
  • Third term: $2 \cdot [(-1)(3) - (0)(-2)] = 2 \cdot [-3 - 0] = 2 \cdot (-3) = -6$

Step 4: Algebraic Summation

$\Delta = 0 + 6 + (-6)$

$\Delta = 6 - 6$

$\Delta = 0$

Matrix Structure Representation: 0 1 2 -1 0 -3 -2 3 0 Result = 0

Final Solution: The value of the determinant is $0$.


More Questions from Class 12 Mathematics Determinants EXERCISE 4.1


CBSE Solutions for Class 12 Mathematics Determinants


Chapters in CBSE - Class 12 Mathematics


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